Question

A solution was made by dissolving 0.7767 g Ba(OH)2 in water to give a final volume of 0.1769 L. What is the molar concentration of OH- in the solution, in M?

a) 4.39 M
b) 2.20 M
c) 8.78 M
d) 1.10 M

Answer 1

The molar concentration of OH- in the solution, found by calculating the moles of Ba(OH)2 and accounting for its dissociation into two OH- ions, is 0.10242 M, which does not match any of the provided options.

Step-by-step explanation:

To find the molar concentration of OH- in the solution, we first calculate the number of moles of Ba(OH)2 dissolved. Since Ba(OH)2 dissociates into one Ba2+ ion and two OH- ions:

• Molecular weight of Ba(OH)2 = 137.33 g/mol (Ba) + 2(16.00 g/mol (O) + 1.01 g/mol (H)) = 171.35 g/mol.
• Number of moles of Ba(OH)2 = mass (g) / molecular weight (g/mol) = 0.7767 g / 171.35 g/mol = 4.53 x 10-3 mol.
• Since each mole of Ba(OH)2 produces two moles of OH-, the moles of OH- = 2 x 4.53 x 10-3 mol = 9.06 x 10-3 mol.

The molar concentration of OH- (Molarity) is then:

Molarity of OH- = moles of OH- / volume of solution (L) = 9.06 x 10-3 mol / 0.1769 L = 0.05121 M.

Since we need to find the concentration in mol/L and one mole of Ba(OH)2 produces two moles of OH-, the final concentration of OH- in the solution is:

Final molar concentration of OH- = 2 x 0.05121 M = 0.10242 M.

Therefore, none of the provided options (a) 4.39 M, (b) 2.20 M, (c) 8.78 M, or (d) 1.10 M match the calculated concentration. The closest option to our calculated concentration is (d) 1.10 M, but it is not the correct answer. There may be an error in the question or the provided options.